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feat: add solutions to lc problems: No.0162,0163 #1771

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Oct 9, 2023
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feat: add solutions to lc problems: No.0162,0163 #1771

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65 changes: 36 additions & 29 deletions solution/0100-0199/0162.Find Peak Element/README.md
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Expand Up @@ -46,7 +46,15 @@

<!-- 这里可写通用的实现逻辑 -->

二分查找。
**方法一：二分查找**

我们定义二分查找的左边界 $left=0$，右边界 $right=n-1$，其中 $n$ 是数组的长度。在每一步二分查找中，我们找到当前区间的中间元素 $mid$，然后比较 $mid$ 与其右边元素 $mid+1$ 的值：

- 如果 $mid$ 的值大于 $mid+1$ 的值，则左侧存在峰值元素，我们将右边界 $right$ 更新为 $mid$；
- 否则，右侧存在峰值元素，我们将左边界 $left$ 更新为 $mid+1$。
- 最后，当左边界 $left$ 与右边界 $right$ 相等时，我们就找到了数组的峰值元素。

时间复杂度 $O(\log n)$，其中 $n$ 是数组 $nums$ 的长度。每一步二分查找可以将搜索区间减少一半，因此时间复杂度为 $O(\log n)$。空间复杂度 $O(1)$。

<!-- tabs:start -->

Expand Down Expand Up @@ -88,22 +96,24 @@ class Solution {
}
```

### **TypeScript**
### **C++**

```ts
function findPeakElement(nums: number[]): number {
let left = 0,
right = nums.length - 1;
while (left < right) {
let mid: number = (left + right) >> 1;
if (nums[mid] <= nums[mid + 1]) {
left = mid + 1;
} else {
right = mid;
```cpp
class Solution {
public:
int findPeakElement(vector<int>& nums) {
int left = 0, right = nums.size() - 1;
while (left < right) {
int mid = left + right >> 1;
if (nums[mid] > nums[mid + 1]) {
right = mid;
} else {
left = mid + 1;
}
}
return left;
}
return left;
}
};
```

### **Go**
Expand All @@ -123,24 +133,21 @@ func findPeakElement(nums []int) int {
}
```

### **C++**
### **TypeScript**

```cpp
class Solution {
public:
int findPeakElement(vector<int>& nums) {
int left = 0, right = nums.size() - 1;
while (left < right) {
int mid = left + right >> 1;
if (nums[mid] > nums[mid + 1]) {
right = mid;
} else {
left = mid + 1;
}
```ts
function findPeakElement(nums: number[]): number {
let [left, right] = [0, nums.length - 1];
while (left < right) {
const mid = (left + right) >> 1;
if (nums[mid] > nums[mid + 1]) {
right = mid;
} else {
left = mid + 1;
}
return left;
}
};
return left;
}
```

### **...**
Expand Down
65 changes: 36 additions & 29 deletions solution/0100-0199/0162.Find Peak Element/README_EN.md
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Expand Up @@ -38,7 +38,15 @@

## Solutions

Binary search.
**Solution 1: Binary Search**

We define the left boundary of binary search as $left=0$ and the right boundary as $right=n-1$, where $n$ is the length of the array. In each step of binary search, we find the middle element $mid$ of the current interval, and compare the values of $mid$ and its right neighbor $mid+1$:

- If the value of $mid$ is greater than the value of $mid+1$, there exists a peak element on the left side, and we update the right boundary $right$ to $mid$.
- Otherwise, there exists a peak element on the right side, and we update the left boundary $left$ to $mid+1$.
- Finally, when the left boundary $left$ is equal to the right boundary $right$, we have found the peak element of the array.

The time complexity is $O(\log n)$, where $n$ is the length of the array $nums$. Each step of binary search can reduce the search interval by half, so the time complexity is $O(\log n)$. The space complexity is $O(1)$.

<!-- tabs:start -->

Expand Down Expand Up @@ -76,22 +84,24 @@ class Solution {
}
```

### **TypeScript**
### **C++**

```ts
function findPeakElement(nums: number[]): number {
let left = 0,
right = nums.length - 1;
while (left < right) {
let mid: number = (left + right) >> 1;
if (nums[mid] <= nums[mid + 1]) {
left = mid + 1;
} else {
right = mid;
```cpp
class Solution {
public:
int findPeakElement(vector<int>& nums) {
int left = 0, right = nums.size() - 1;
while (left < right) {
int mid = left + right >> 1;
if (nums[mid] > nums[mid + 1]) {
right = mid;
} else {
left = mid + 1;
}
}
return left;
}
return left;
}
};
```

### **Go**
Expand All @@ -111,24 +121,21 @@ func findPeakElement(nums []int) int {
}
```

### **C++**
### **TypeScript**

```cpp
class Solution {
public:
int findPeakElement(vector<int>& nums) {
int left = 0, right = nums.size() - 1;
while (left < right) {
int mid = left + right >> 1;
if (nums[mid] > nums[mid + 1]) {
right = mid;
} else {
left = mid + 1;
}
```ts
function findPeakElement(nums: number[]): number {
let [left, right] = [0, nums.length - 1];
while (left < right) {
const mid = (left + right) >> 1;
if (nums[mid] > nums[mid + 1]) {
right = mid;
} else {
left = mid + 1;
}
return left;
}
};
return left;
}
```

### **...**
Expand Down
11 changes: 5 additions & 6 deletions solution/0100-0199/0162.Find Peak Element/Solution.ts
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@@ -1,12 +1,11 @@
function findPeakElement(nums: number[]): number {
let left = 0,
right = nums.length - 1;
let [left, right] = [0, nums.length - 1];
while (left < right) {
let mid: number = (left + right) >> 1;
if (nums[mid] <= nums[mid + 1]) {
left = mid + 1;
} else {
const mid = (left + right) >> 1;
if (nums[mid] > nums[mid + 1]) {
right = mid;
} else {
left = mid + 1;
}
}
return left;
Expand Down
106 changes: 55 additions & 51 deletions solution/0100-0199/0163.Missing Ranges/README.md
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Expand Up @@ -48,9 +48,9 @@

**方法一：模拟**

按照题意模拟即可。
我们直接按照题意模拟即可。

时间复杂度 $O(n)$，忽略答案的空间消耗，空间复杂度 $O(1)$。其中 $n$ 为数组 `nums` 的长度。
时间复杂度 $O(n)$，其中 $n$ 为数组 $nums$ 的长度。忽略答案的空间消耗，空间复杂度 $O(1)$。

<!-- tabs:start -->

Expand All @@ -60,21 +60,20 @@

```python
class Solution:
def findMissingRanges(self, nums: List[int], lower: int, upper: int) -> List[str]:
def f(a, b):
return str(a) if a == b else f'{a}->{b}'

def findMissingRanges(
self, nums: List[int], lower: int, upper: int
) -> List[List[int]]:
n = len(nums)
if n == 0:
return [f(lower, upper)]
return [[lower, upper]]
ans = []
if nums[0] > lower:
ans.append(f(lower, nums[0] - 1))
ans.append([lower, nums[0] - 1])
for a, b in pairwise(nums):
if b - a > 1:
ans.append(f(a + 1, b - 1))
ans.append([a + 1, b - 1])
if nums[-1] < upper:
ans.append(f(nums[-1] + 1, upper))
ans.append([nums[-1] + 1, upper])
return ans
```

Expand All @@ -84,31 +83,25 @@ class Solution:

```java
class Solution {
public List<String> findMissingRanges(int[] nums, int lower, int upper) {
public List<List<Integer>> findMissingRanges(int[] nums, int lower, int upper) {
int n = nums.length;
List<String> ans = new ArrayList<>();
if (n == 0) {
ans.add(f(lower, upper));
return ans;
return List.of(List.of(lower, upper));
}
List<List<Integer>> ans = new ArrayList<>();
if (nums[0] > lower) {
ans.add(f(lower, nums[0] - 1));
ans.add(List.of(lower, nums[0] - 1));
}
for (int i = 1; i < n; ++i) {
int a = nums[i - 1], b = nums[i];
if (b - a > 1) {
ans.add(f(a + 1, b - 1));
if (nums[i] - nums[i - 1] > 1) {
ans.add(List.of(nums[i - 1] + 1, nums[i] - 1));
}
}
if (nums[n - 1] < upper) {
ans.add(f(nums[n - 1] + 1, upper));
ans.add(List.of(nums[n - 1] + 1, upper));
}
return ans;
}

private String f(int a, int b) {
return a == b ? a + "" : a + "->" + b;
}
}
```

Expand All @@ -117,27 +110,22 @@ class Solution {
```cpp
class Solution {
public:
vector<string> findMissingRanges(vector<int>& nums, int lower, int upper) {
auto f = [](int a, int b) {
return a == b ? to_string(a) : to_string(a) + "->" + to_string(b);
};
vector<vector<int>> findMissingRanges(vector<int>& nums, int lower, int upper) {
int n = nums.size();
vector<string> ans;
if (n == 0) {
ans.emplace_back(f(lower, upper));
return ans;
return {{lower, upper}};
}
vector<vector<int>> ans;
if (nums[0] > lower) {
ans.emplace_back(f(lower, nums[0] - 1));
ans.push_back({lower, nums[0] - 1});
}
for (int i = 1; i < n; ++i) {
int a = nums[i - 1], b = nums[i];
if (b - a > 1) {
ans.emplace_back(f(a + 1, b - 1));
for (int i = 1; i < nums.size(); ++i) {
if (nums[i] - nums[i - 1] > 1) {
ans.push_back({nums[i - 1] + 1, nums[i] - 1});
}
}
if (nums[n - 1] < upper) {
ans.emplace_back(f(nums[n - 1] + 1, upper));
ans.push_back({nums[n - 1] + 1, upper});
}
return ans;
}
Expand All @@ -147,34 +135,50 @@ public:
### **Go**

```go
func findMissingRanges(nums []int, lower int, upper int) (ans []string) {
f := func(a, b int) string {
if a == b {
return strconv.Itoa(a)
}
return strconv.Itoa(a) + "->" + strconv.Itoa(b)
}
func findMissingRanges(nums []int, lower int, upper int) (ans [][]int) {
n := len(nums)
if n == 0 {
ans = append(ans, f(lower, upper))
return
return [][]int{{lower, upper}}
}
if nums[0] > lower {
ans = append(ans, f(lower, nums[0]-1))
ans = append(ans, []int{lower, nums[0] - 1})
}
for i := 1; i < n; i++ {
a, b := nums[i-1], nums[i]
if b-a > 1 {
ans = append(ans, f(a+1, b-1))
for i, b := range nums[1:] {
if a := nums[i]; b-a > 1 {
ans = append(ans, []int{a + 1, b - 1})
}
}
if nums[n-1] < upper {
ans = append(ans, f(nums[n-1]+1, upper))
ans = append(ans, []int{nums[n-1] + 1, upper})
}
return
}
```

### **TypeScript**

```ts
function findMissingRanges(nums: number[], lower: number, upper: number): number[][] {
const n = nums.length;
if (n === 0) {
return [[lower, upper]];
}
const ans: number[][] = [];
if (nums[0] > lower) {
ans.push([lower, nums[0] - 1]);
}
for (let i = 1; i < n; ++i) {
if (nums[i] - nums[i - 1] > 1) {
ans.push([nums[i - 1] + 1, nums[i] - 1]);
}
}
if (nums[n - 1] < upper) {
ans.push([nums[n - 1] + 1, upper]);
}
return ans;
}
```

### **...**

```
Expand Down
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